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The length of a rectangle is 15 and its width is w.
The perimeter of the rectangle is, at most, 50,
Which inequality can be used to find the longes
possible width?
A
30+ 2x < 50
B. 30 + 2w < 50
D. 30+ 2w > 50
C.30 + 2w > 50

Respuesta :

A rectangle with sides 15 and w has perimeter

[tex]2p = 2(15+w) = 2w+30[/tex]

We want this quantity to be at most 50, so it must be less than or equal to 50:

[tex]2w+30\leq 50[/tex]

For the record, this implies that

[tex]2w\leq 20 \iff w \leq 10[/tex]

So, the width can be at most 10.

ghanami

Answer:

2x+30< 50   ...answer : A

Step-by-step explanation:

the perimeter is :  P = 2(L+W)

let : L= x   given : L=15

P= 2(x+15)

The perimeter of the rectangle is, at most, 50:   P<50

2(x+15) < 50

2x+30< 50   ...answer : A

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